Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I understand that this question would be trivial for experts, sorry for that, I just need to clarify things.

So let $S(\mathbb{R}^n)$ denote the Schwartz space on $\mathbb{R}^n$ and $W_p$, $W_q$ are the Sobolev spaces, or in the other words completions of $S(\mathbb{R}^n)$ with respect to $p$ and $q$ - Sobolev norms.

Let us assume that $\phi: $ $W_q\rightarrow W_p$ is a map , continuously extending the identity map on $S$ assuming that $p < q$.

Question: why is $\phi$ an embedding? It is only clear that $\phi$ is embedding of $S$ ( as it is identity on it).

Similar question for Sobolev embedding theorem into $C^k$.

P.S. To clarify my question here is what I mean by the Sobolev space. For a real number $p\in\mathbb{R}$ the $p$-Sobolev norm on $S$ is given by ${|f|}^2_p=\int{(1+|\xi|)}^{2p}{|\hat{f}(\xi)|}^2d\xi$, where $\hat{f}(\xi)$ is a Fourier transform of $f(x)\in S$. So $W_p$ is a completion of $S$ with respect to this norm.

P.P.S. To make things little more clear I am mostly interested in case of non-integer or negative $p$ when the Sobolev $p$-norm can not be defined through the weak derivatives.

share|improve this question
    
In the question instead of p one should read $p<q$. This is probably a MO bug as I was not able to type it directly in the question. –  Axel Oct 25 '12 at 6:39
    
I guess by Sobolev space $W^p$ you mean what is (nowadays) usually known as $H^p$ - that is, $W^{p,2}$, the space of $p$-times weakly differentiable $L^2$-functions with weak derivatives in $L^2$. Or else do you mean the first Sobolev space? That is, $W^{1,p}$, the space of (once) weakly differentiable $L^p$-functions with values in $L^p$? –  Delio Mugnolo Oct 25 '12 at 7:55
    
To your postscript: this is what I was assuming in my answer. –  András Bátkai Oct 25 '12 at 8:43
    
Delio, what you said is a particular case of positive integer $p$. I am interested in any $p\in\mathbb{R}$ –  Axel Oct 25 '12 at 8:48
    
Andras, if $p$ is negative or non-integer, then it is not about derivatives, that is the point. –  Axel Oct 25 '12 at 8:50

2 Answers 2

up vote 2 down vote accepted

I would like to expand a bit what Delio said. Your question is a bit confusing as it is, but we may assume that you mean that $p$ and $q$ are the parameters representing the derivatives. Then what you need is $$ \|f\|_p \leq C\|f\|_q.$$

If you write out the definition of the Sobolev norm on $S$, then you see immediately that this holds. This implies that the identity map defined on $S$ extends uniquely and continuously to the whole $H^q$. This gives you the embedding. (Strictly speaking you have to verify that this extension is the identity map, but this follows easily from continuity considerations.)

share|improve this answer
    
Andras, please see my revised original post. The issue is that extending the identity map $S\rightarrow S$ by continuity you have only the map $\phi: W_q\rightarrow W_p$. My question is how do you know that $\phi$ has no kernel. –  Axel Oct 25 '12 at 8:46
    
See my remark in the brackets. You can verify that your operator acts as the identity by hand. Hence, no kernel. –  András Bátkai Oct 25 '12 at 8:49
    
Andras, the thing in brackets which I do not understand is the following: When you say that you have to verify that extension is the identity map, then you mean " identity map" on what space? More precisely $W_p$ and $W_q$ are in general $\textbf{different}$ spaces and the only thing they have " in common" is the space $S$. Now the map $\phi: W_q\rightarrow\W_p$ is a map between different spaces which is an identity when restricted to $S$ ( under natural identification of $S$ in $W_q$ and $S$ in $W_p$.) But( to be continued) –  Axel Oct 25 '12 at 9:01
    
..( continue) before you establish that $\phi$ has no kernel you can not think of $W_q$ as a subspace of $W_p$ so you can not talk about "identity map" on $W_q$. –  Axel Oct 25 '12 at 9:02
1  
Sorry, I was away. Because your original sequence has a subsequence converging pointwise almost everywhere. –  András Bátkai Oct 25 '12 at 14:07

Axel,

please read my above comment. If by $W^p$ you mean $W^{p,2}$, then the imbedding follows from the very definition of Sobolev space - or rather, if you wish to keep your definition, from the Meyer-Serrin theorem which essentially establishes the equivalence between your definition and mine.

If by $W^p$ you mean $W^{1,p}$, then $\phi$ is an embedding only if you have a finite measure space (and in this case the assertion follows from the embedding of $L^q$ ind $L^p$ by Hölder's inequality), otherwise it is in general wrong (think of the Sobolev spaces over $\mathbb R^n$).

share|improve this answer
    
Delio, please see my revised original post. –  Axel Oct 25 '12 at 8:43
    
Good, then my first answer is correct. It is well-known that your definition based on the Fourier transform is equivalent to the other both ones, see en.wikipedia.org/wiki/Sobolev_space#Bessel_potential_spaces to begin with. –  Delio Mugnolo Oct 25 '12 at 9:31
    
Delio, in the wikipedia link that you gave it is explicitly assumed that $p$ is a positive integer which is not the case of interest to me. –  Axel Oct 25 '12 at 9:46
    
Sorry, I have overseen that. A possibility (not necessarily the easiest one, though) is then to observe that Sobolev spaces of real order can then be equivalently defined by means of complex interpolation (check wikipedia, again, or Lunardi's monograph on interpolation theory) and then observe that if $X$ is a subspace of $Y$ you have the wished inclusion relation for two interpolation spaces $[X,Y]_{\theta_1},[X,Y]_{\theta_2}$. –  Delio Mugnolo Oct 25 '12 at 12:27
    
Thank you for your comment Delio, I think I will try to find somehow easier explanation, only because I do not have enough time resourses to check in detail the deeper explanations which you mention. I am quite surprised that this question requires so advanced tools as I truly believed this is something well-known. –  Axel Oct 26 '12 at 8:43

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.